Question

# Two different analytical tests can be used to determine the impurity level in steel alloys. Nine specimens are tested using both procedures, and the results are shown in the tabulation.Find the 95\% confidence interval for the mean difference of the tests, and explain it.

Confidence intervals

Two different analytical tests can be used to determine the impurity level in steel alloys. Nine specimens are tested using both procedures, and the results are shown in the following tabulation.
$$\begin{array}{|c|c|}\hline Specimen & Test-1 & Test-2 \\ \hline 1 & 1.2 & 1.4 \\ \hline 2 & 1.6 & 1.7 \\ \hline 3 & 1.5 & 1.5 \\ \hline 4 & 1.4 & 1.3 \\ \hline 5 & 1.8 & 2.0 \\ \hline 6 & 1.8 & 2.1 \\ \hline 7 & 1.4 & 1.7 \\ \hline 8 & 1.5 & 1.6 \\ \hline 9 & 1.4 & 1.5 \\ \hline \end{array}$$
Find the $$\displaystyle{95}\%$$ confidence interval for the mean difference of the tests, and explain it.

## Expert Answers (1)

2021-08-03

Step 1
$$\begin{array}{|c|c|}\hline Test-1(X) & Test-2(Y) & (x-\bar{x}) & (x-\bar{x})^{2} & (y-\bar{y}) & (y-bar{y})^{2} \\ \hline 1.2 & 1.4 & -0.31111 & 0.09679 & -0.24444 & 0.05975 \\ \hline 1.6 & 1.7 & 0.08889 & 0.00790 & 0.05556 & 0.00309 \\ \hline 1.5 & 1.5 & -0.01111 & 0.00012 & -0.14444 & 0.02086 \\ \hline 1.4 & 1.3 & -0.11111 & 0.01235 & -0.34444 & 0.11864 \\ \hline 1.8 & 2.0 & 0.28889 & 0.08346 & 0.35556 & 0.12642 \\ \hline 1.8 & 2.1 & 0.28889 & 0.08346 & 0.45556 & 0.20753 \\ \hline 1.4 & 1.7 & -0.11111 & 0.01235 & 0.05556 & 0.00309 \\ \hline 1.5 & 1.6 & -0.01111 & 0.00012 & -0.04444 & 0.00198\\ \hline 1.4 & 1.5 & -0.11111 & 0.01235 & -0.14444 & 0.02086\\ \hline \end{array}$$
Step 2
The mean is given by
$$\displaystyle\overline{{{x}}}={\frac{{{1}}}{{{n}}}}\sum{x}$$
Test 1 NKS $$\displaystyle\overline{{{x}}}={\frac{{{13.6}}}{{{9}}}}={1.511}$$
Test 2
$$\displaystyle\overline{{{y}}}={\frac{{{14.8}}}{{{9}}}}={1.644}$$
The standard deviation is given by
$$\displaystyle\sigma=\sqrt{{{\frac{{{1}}}{{{n}-{1}}}}\sum{\left({x}-\overline{{{x}}}\right)}^{{{2}}}}}$$
Test 1
$$\displaystyle\sigma=\sqrt{{{\frac{{{0.30889}}}{{{8}}}}}}={0.1965}$$
Test 2
$$\displaystyle{s}_{{{2}}}=\sqrt{{{\frac{{{0.5622}}}{{{8}}}}}}={0.2651}$$
The $$\displaystyle{95}\%$$ CI is given by
$$\displaystyle{C}.{I}.={\left(\overline{{{x}}}-\overline{{{y}}}\right)}\pm{t}_{{{0.05},\ {8}}}\sqrt{{{\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{n}_{{{1}}}}}}+{\frac{{{{s}_{{{2}}}^{{{2}}}}}}{{{n}_{{{2}}}}}}}}$$
$$\displaystyle{C}.{I}.={\left({1.511}-{1.644}\right)}\pm{2.306}\sqrt{{{\frac{{{\left({0.1965}\right)}^{{{2}}}}}{{{9}}}}+{\frac{{{\left({0.2651}\right)}^{{{2}}}}}{{{9}}}}}}$$
$$\displaystyle{C}.{I}.={\left[-{0.387},\ {0.120}\right]}$$
It shows that the true difference between the mean of test 1 and test 2 will lie within -0.387 and 0.120.